Introduction To Health Physics
5th Edition
ISBN: 9780071835275
Author: Johnson, Thomas E. (thomas Edward), Cember, Herman.
Publisher: Mcgraw-hill Education,
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Chapter 9, Problem 9.26P
(a)
To determine
The value of the series resistor
(b)
To determine
Upper limit of the output voltage pulse
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The linear attenuation coefficient for 2.0-MeV gamma rays in water is 4.9 m-1 and 52 m-1 in lead. What thickness of water would give the same shielding for gamma rays as 15 mm of lead?
Ionizing radiation enters a Geiger tube with 1.45 MeV of energy. As the radiation passes through the tube, all of this energy goes into
creating ion pairs and each ion pair requires 30 eV of energy.
(a) If an applied voltage sweeps these ions out of the gas in 1.10 µs, determine the current (in A).
1.19e-8
X
What is the definition of current? How many electronic charges constitute an ion pair? How can you determine the number of ion
pairs created? A
(b) Since the applied voltage in the Geiger tube accelerates these separated ions, creating other ion pairs in subsequent collisions, the
actual current is greater than that determined in part (a). If this effect multiplies the number of ion pairs by 880, determine the
actual current (in A).
1.05e-6
X
How is the increase in current related to the increase in number of ion pairs? A
Suppose a particle of ionizing radiation deposits 1.0 MeV in the gas of a Geiger tube, all of which goes to creating ion pairs. Each ion pair requires 30.0 eV of energy. (a) The applied voltage sweeps the ions out of the gas in 1.00 µs. What is the current? (b) This current is smaller than the actual current since the applied voltage in the Geiger tube accelerates the separated ions, which then create other ion pairs in subsequent collisions. What is the current if this last effect multiplies the number of ion pairs by 900?
Chapter 9 Solutions
Introduction To Health Physics
Ch. 9 - Prob. 9.1PCh. 9 - Prob. 9.2PCh. 9 - Prob. 9.3PCh. 9 - Prob. 9.4PCh. 9 - Prob. 9.5PCh. 9 - Prob. 9.6PCh. 9 - Prob. 9.7PCh. 9 - Prob. 9.8PCh. 9 - Prob. 9.9PCh. 9 - Prob. 9.10P
Ch. 9 - Prob. 9.11PCh. 9 - Prob. 9.12PCh. 9 - Prob. 9.13PCh. 9 - A counting system has a background of 360 counts...Ch. 9 - Prob. 9.15PCh. 9 - Prob. 9.16PCh. 9 - Prob. 9.17PCh. 9 - Prob. 9.18PCh. 9 - Prob. 9.19PCh. 9 - Prob. 9.20PCh. 9 - Prob. 9.21PCh. 9 - Prob. 9.22PCh. 9 - Prob. 9.23PCh. 9 - Prob. 9.24PCh. 9 - Prob. 9.25PCh. 9 - Prob. 9.26PCh. 9 - Prob. 9.27PCh. 9 - Prob. 9.28PCh. 9 - Prob. 9.29PCh. 9 - Prob. 9.31PCh. 9 - Prob. 9.33PCh. 9 - Prob. 9.34PCh. 9 - Prob. 9.35PCh. 9 - Prob. 9.38PCh. 9 - Prob. 9.39PCh. 9 - Prob. 9.44P
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